Amit’s bank charges

The math club needs to raise more than $552. $.50 for a trip to state competition. The club has raise 12% of the funds which is

ratelena [41]

Each of the members must raise at least $69.46.

To start they need to raise $552.50. However, they have already raised 12% of it. They only need to raise 88% of it.

552.50 x 0.88 = 486.2

Divide the remaining amount by 7 to determine how much each individual must raise.

486.2 / 7 = 69.46

4 0

3 years ago

Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one anothe

scZoUnD [109]

Answer:

a) P(male=blue or female=blue) = 0.71

b) P(female=blue | male=blue) = 0.68

c) P(female=blue | male=brown) = 0.35

d) P(female=blue | male=green) = 0.31

e) We can conclude that the eye colors of male respondents and their partners are not independent.

Step-by-step explanation:

We are given following information about eye colors of 204 Scandinavian men and their female partners.

Blue Brown Green Total

Blue 78 23 13 114

Brown 19 23 12 54

Green 11 9 16 36

Total 108 55 41 204

a) What is the probability that a randomly chosen male respondent or his partner has blue eyes?

Using the addition rule of probability,

∵ P(A or B) = P(A) + P(B) - P(A and B)

For the given case,

P(male=blue or female=blue) = P(male=blue) + P(female=blue) - P(male=blue and female=blue)

P(male=blue or female=blue) = 114/204 + 108/204 − 78/204

P(male=blue or female=blue) = 0.71

b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=blue) = 78/114

P(female=blue | male=blue) = 0.68

c) What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=brown) = 19/54

P(female=blue | male=brown) = 0.35

d) What is the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=green) = 11/36

P(female=blue | male=green) = 0.31

e) Does it appear that the eye colors of male respondents and their partners are independent? Explain

If the following relation holds true then we can conclude that the eye colors of male respondents and their partners are independent.

∵ P(B | A) = P(B)

P(female=blue | male=brown) = P(female=blue)

or alternatively, you can also test

P(female=blue | male=green) = P(female=blue)

P(female=blue | male=blue) = P(female=blue)

But

P(female=blue | male=brown) ≠ P(female=blue)

19/54 ≠ 108/204

0.35 ≠ 0.53

Therefore, we can conclude that the eye colors of male respondents and their partners are not independent.

7 0

3 years ago

What is the equation of the graphed linear model?

ozzi

Hello,
The line passes through (-1,3) and (9,11)

slope=(11-3)/(9+1)=8/10=4/5

y-3=4/5(x+1)
==>y=4/5 *x +19/5

3 0

3 years ago

Can someone actually help me please?

Tamiku [17]

D and B is the anwser

5 0

2 years ago

The principal at Crest Middle School, which enrolls only sixth-grade students and seventh-grade students, is interested in deter

AlekseyPX

Answer:

a) [ -27.208 , -12.192 ]

b) New procedure is not recommended

Step-by-step explanation:

Solution:-

- It is much more common for a statistical analyst to be interested in the difference between means than in the specific values of the means themselves.

- The principal at Crest Middle School collects data on how much time students at that school spend on homework each night.

- He/She takes a " random " sample of n = 20 from a sixth and seventh grades students from the school population to conduct a statistical analysis.

- The summary of sample mean ( x1 & x2 ) and sample standard deviation ( s1 & s2 ) of the amount of time spent on homework each night (in minutes) for each grade of students is given below:

<u>Mean ( xi )</u> <u> Standard deviation ( si )</u>

Sixth grade students 27.3 10.8

Seventh grade students 47.0 12.4

- We will first check the normality of sample distributions.

We see that sample are "randomly" selected.
The mean times are independent for each group
The groups are selected independent " sixth " and " seventh" grades.

The means of both groups are conforms to 10% condition of normality.

Hence, we will assume that the samples are normally distributed.

- We are to construct a 95% confidence interval for the difference in means ( u1 and u2 ).

- Under the assumption of normality we have the following assumptions for difference in mean of independent populations:

Population mean of 6th grade ( u1 ) ≈ sample mean of 6th grade ( x1 )

Population mean of 7th grade ( u2 ) ≈ sample mean of 6th grade ( x2 )

Therefore, the difference in population mean has the following mean ( u ^ ):

u^ = u1 - u2 = x1 - x2

u^ = 27.3 - 47.0

u^ = -19.7

- Similarly, we will estimate the standard deviation (Standard Error) for a population ( σ^ ) represented by difference in mean. The appropriate relation for point estimation of standard deviation of difference in means is given below:

σ^ = √ [ ( σ1 ^2 / n1 ) + ( σ2 ^2 / n2 ) ]

Where,

σ1 ^2 : The population variance for sixth grade student.

σ2 ^2 : The population variance for sixth grade student.

n1 = n2 = n : The sample size taken from both populations.

Therefore,

σ^ = √ [ ( 2*σ1 ^2 / n )].

- Here we will assume equal population variances : σ1 ≈ σ2 ≈ σ is "unknown". We can reasonably assume the variation in students in general for the different grade remains somewhat constant owing to other reasons and the same pattern is observed across.

- The estimated standard deviation ( σ^ ) of difference in means is given by:

σ^ =

$s_p*\sqrt{\frac{1}{n_1} + \frac{1}{n_2} } = s_p*\sqrt{\frac{1}{n} + \frac{1}{n} } = s_p*\sqrt{\frac{2}{n}}\\\\\\s_p = \sqrt{\frac{(n_1 - 1 )*s_1^2 + (n_2 - 1 )*s_2^2}{n_1+n_2-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{n+n-2} } = \sqrt{\frac{(n - 1 )*s_1^2 + (n - 1 )*s_2^2}{2n-2} } \\\\s_p = \sqrt{\frac{(20 - 1 )*s_1^2 + (20 - 1 )*s_2^2}{2(20)-2} } \\\\s_p = \sqrt{\frac{19*10.8^2 + 19*12.4^2}{38} } = \sqrt{135.2} \\\\s_p = 11.62755$

σ^ = 11.62755*√2/20

σ^ = 3.67695

- Now we will determine the critical value associated with Confidence interval ( CI ) which is defined by the standard probability of significance level ( α ). Such that:

Significance Level ( α ) = 1 - CI = 1 - 0.95 = 0.05

- The reasonable distribution ( T or Z ) would be determined on the basis of following conditions:

The population variances ( σ1 ≈ σ2 ≈ σ ) are unknown.
The sample sizes ( n1 & n2 ) are < 30.

Hence, the above two conditions specify the use of T distribution critical value. The degree of freedom ( v ) for the given statistics is given by:

v = n1 + n2 - 2 = 2n - 2 = 2*20 - 2

v = 38 degrees of freedom

- The t-critical value is defined by the half of significance level ( α / 2 ) and degree of freedom ( v ) as follows:

t-critical = t_α / 2, v = t_0.025,38 = 2.024

- Then construct the interval for 95% confidence as follows:

[ u^ - t-critical*σ^ , u^ + t-critical*σ^ ]

[ -19.7 - 2.042*3.67695 , -19.7 + 2.042*3.67695 ]

[ -19.7 - 7.5083319 , -19.7 + 7.5083319 ]

[ -27.208 , -12.192 ]

- The principal should be 95% confident that the difference in mean times spent of homework for ALL 6th and 7th grade students in this school (population) lies between: [ -27.208 , -12.192 ]

- The procedure that the matched-pairs confidence interval for the mean difference in time spent on homework prescribes the integration of time across different sample groups.

- If we integrate the times of students of different grades we would have to make further assumptions like:

The intelligence levels of different grade students are same
The aptitude of students from different grades are the same
The efficiency of different grades are the same.

- We have to see that both samples are inherently different and must be treated as separate independent groups. Therefore, the above added assumptions are not justified to be used for the given statistics. The procedure would be more bias; hence, not recommended.

8 0

3 years ago